Applications of Similarity: Practice

Right base angles and equal parallel-ray tip angles give AA. Match height with height and own shadow with own shadow.

$\begin{aligned}H&=12.0\cdot\frac{1.25}{0.750},\\&=20.0\text{ m}.\end{aligned}$

AA pairs the two heights and their own ground shadows. The staff is one fifth of the object height.

$\begin{aligned}s&=6.40\cdot\frac{1.80}{9.00},\\&=1.28\text{ m}.\end{aligned}$

The right angles and equal mirror angles give AA. Object height over eye height equals object ground leg over observer ground leg.

$\begin{aligned}H&=1.50\cdot\frac{14.0}{2.5},\\&=8.4\text{ m}.\end{aligned}$

AA matches vertical heights and ground legs. Keep the observer triangle first in both ratios.

$\begin{aligned}d&=24.0\cdot\frac{1.65}{18.0},\\&=2.2\text{ m}.\end{aligned}$

Right angles at A,C and vertical angles at B give AA. AT matches CD and AB matches CB.

$\begin{aligned}AT&=10.2\cdot\frac{30.0}{18.0},\\&=17.0\text{ m}.\end{aligned}$

Apply the scale in common centimeter units, then convert.

Convert the real length to centimeters before dividing.

Adding the corresponding scaled edges multiplies the whole boundary sum by 1.75.

$\begin{aligned}P_{\rm image}&=1.75\cdot 28,\\&=49\text{ cm}.\end{aligned}$

The scale factor from the original to the image is eleven eighths and applies to all matched boundary sides.

$\begin{aligned}P_{\rm image}&=40\cdot\frac{11}{8},\\&=55\text{ cm}.\end{aligned}$

Convert the real length to millimeters, then map from real to drawing.

Add all route legs, rather than measuring the direct endpoint gap.

Every printed length is eight tenths of the old drawing length while real lengths stay fixed.

$\begin{aligned}\frac{L_{\rm print}}{L_{\rm real}}&=\frac{0.8}{2000},\\&=\frac 1{2500}.\end{aligned}$